C3H18(1) + 2502(g) - 16CO2(9) + 18H20(1) AH°n = -11,020 KJ/mol. iven that AH°{CO2(9)] = -393.5 kJ/mol and AH?{H2O(1)] = -285.8 kJ/mol, calculate the standard enthalpy of formatio O -420 kJ/mol O 22,040 kJ/mol O -210 kJ/mol O 420 kJ/mol O -11,230 kJ/mol
C3H18(1) + 2502(g) - 16CO2(9) + 18H20(1) AH°n = -11,020 KJ/mol. iven that AH°{CO2(9)] = -393.5 kJ/mol and AH?{H2O(1)] = -285.8 kJ/mol, calculate the standard enthalpy of formatio O -420 kJ/mol O 22,040 kJ/mol O -210 kJ/mol O 420 kJ/mol O -11,230 kJ/mol
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![Octane (C3H18) undergoes combustion according to the following thermochemical equation:
2C3H18(1) + 2502(g)
16CO2(g) + 18H20(1)
°pxn = -11,020 kJ/mol.
Given that AH°{CO2(g)] = -393.5 kJ/mol and AH°{{H2O(1)] = -285.8 kJ/mol, calculate the standard enthalpy of formation of octane.
O -420 kJ/mol
O 22,040 kJ/mol
O - 210 kJ/mol
O 420 kJ/mol
O -11,230 kJ/mol](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4185167c-97ef-450e-81af-f379fc668246%2F7038aa90-3928-4205-80d4-4a3cc32d9a00%2Fsu6s3zp_processed.png&w=3840&q=75)
Transcribed Image Text:Octane (C3H18) undergoes combustion according to the following thermochemical equation:
2C3H18(1) + 2502(g)
16CO2(g) + 18H20(1)
°pxn = -11,020 kJ/mol.
Given that AH°{CO2(g)] = -393.5 kJ/mol and AH°{{H2O(1)] = -285.8 kJ/mol, calculate the standard enthalpy of formation of octane.
O -420 kJ/mol
O 22,040 kJ/mol
O - 210 kJ/mol
O 420 kJ/mol
O -11,230 kJ/mol
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